Question

If \( y \frac{dy}{dx} = x \left[ \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} + \frac{y^2}{x^2} \right], x>0, \phi>0, \) and \( y(1) = -1 \), then \( \phi\left(\frac{y^2}{4}\right) \) is equal to :

• A. \( \phi(1) \)
• B. \( 2\phi(1) \)
• C. \( 4\phi(1) \)
• D. \( 4\phi(2) \)

Hint:

Whenever a differential equation contains terms like \( \phi(y/x) \) or \( \phi(y^2/x^2) \), it is likely homogeneous. Substitution \( v = y/x \) or \( v = y^2/x^2 \) should be your first step.

Answer 1

The Correct Option is C

To solve the given differential equation problem: 

The equation provided is \( y \frac{dy}{dx} = x \left( \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} + \frac{y^2}{x^2} \right) \). We need to find the expression for \( \phi(y^2/4) \) given \( y(1) = -1 \).

1. Change of Variables: Let us simplify the equation by using a suitable substitution. Let \( v = \frac{y}{x} \), which implies \( y = vx \).
2. Then, differentiating both sides with respect to \( x \), we have \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
3. Substitute into the original differential equation:

\(y \left( v + x \frac{dv}{dx} \right) = x \left( \frac{\phi(v^2)}{\phi'(v^2)} + v^2 \right)\)

1. Replace \( y \) with \( vx \):

\(vx (v + x \frac{dv}{dx}) = x \left( \frac{\phi(v^2)}{\phi'(v^2)} + v^2 \right)\)

1. Cancel out \( x \) from both sides:

\(v (v + x \frac{dv}{dx}) = \frac{\phi(v^2)}{\phi'(v^2)} + v^2\)

1. Rearrange the equation:

\(x \frac{dv}{dx} = \frac{\phi(v^2)}{\phi'(v^2)}\)

1. This is a separable differential equation. Separate variables to integrate:

\(\frac{\phi'(v^2) dv}{\phi(v^2)} = \frac{dx}{x}\)

1. Integrate both sides:

\(\ln(\phi(v^2)) = \ln(x) + C\)

1. Remove the logarithm by exponentiating both sides:

\(\phi(v^2) = kx\) where \( k = e^C \).

1. Use the initial condition \( y(1) = -1 \), which gives \( v = \frac{y}{x} = -1 \) when \( x = 1 \).
2. Substitute \( v = -1 \) and \( x = 1 \) into the expression:

\(\phi((-1)^2) = k\) hence \( \phi(1) = k \).

1. Now, substitute into the equation for which we need to find \( \phi(\frac{y^2}{4}) \): Since. \(< y = vx \Rightarrow > y^2 = v^2 x^2 \) set, \(x = 2\) when \( y^2/4 = v^2 \), gives \(v^2 x^2 = 4v^2, \phi(4v^2) = 4\phi(1)\)

Therefore, the value of \( \phi\left(\frac{y^2}{4}\right) \) is \( 4\phi(1) \).